Problem: For a function $h$, we are given that $h(5)=2$ and $h'(5)=-6$. What's the equation of the tangent line to the graph of $h$ at $x=5$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-5=2(x+6)$ (Choice B) B $y+6=2(x-5)$ (Choice C) C $y-5=-6(x-2)$ (Choice D) D $y-2=-6(x-5)$
The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $h'(5)$ gives the slope of the tangent line to the graph of $h$ where $x=5$. We are given that $h'(5)=-6$, so the slope of the tangent line is $-6$. Furthermore, we are given that $h(5)=2$, which means the point of intersection of the tangent line and the graph is $(5,2)$. To summarize, the tangent line has a slope of $-6$ and it passes through the point $(5,2)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-2&=-6(x-5) \\\\ \end{aligned}$ The equation is $y-2=-6(x-5)$.